I Two-sided Prinicipal Ideal - the Noncommutative Case - D&F

I have a basic question regarding the generation of a two sided principal ideal in the noncommutative case ...

In Section 7.4 on pages 251-252 Dummit and Foote write the following:

In the above text we read:

" ... ... If ##R## is not commutative, however, the set ##\{ ras \ | \ r, s \in R \}## is not necessarily the two-sided ideal generated by ##a## since it need not be closed under addition (in this case the ideal generated by ##a## is the ideal ##RaR##, which consists of all finite sums of elements of the form ##ras, r,s \in R##). ... ... "

I must confess I do not understand or follow this argument ... I hope someone can clarify (slowly and clearly
) what it means ... ...

Staff: Mentor

In an ideal ##\mathcal{I}## you have ##x+y\in \mathcal{I}## for elements ##x\,,\,y\in \mathcal{I}##. This is a basic part of its definition.
If you now have two elements ##ras \; (r,s\in R\,; a\in \mathcal{I})## and ##paq \; (p,q \in R\,; a\in \mathcal{I})## there is - in general - no way to write ##ras+paq=uav## because you cannot pull the factors ##r,s,p,q## on the other side of ##a##.
Therefore ##\mathcal{I}=(a) = RaR = LC(\{ras \,|\, r,s \in R\}) \supsetneq \{ras \,|\, r,s \in R\}## is - in general - a proper inclusion. The latter is only a set.

In the commutative case we have
##ras+paq=r(as)+p(aq)=r(sa)+p(qa)=(rs)a+(pq)a=(rs+pq)a## and all linear combinations are of the form ##ua\;(u=rs+pq \in R)##.
However, this calculation is not allowed in non-commutative rings.